# QFT Hilbert spaces over other rings than the complex numbers $\mathbb{C}$

I would like some help evaluating a physics theory recently proposed by a physics professor at the College of Dupage.

I think the theory is utterly wrong, for very simple reasons. If an amateur feels he can disprove a physics Professor (especially if using simple logic and basic algebra, and especially when involving quantum mechanics), its usually not a good sign to put it mildly. I want to learn here, and need physics help from you experts to evaluate the theory.

I have taken some physics classes back in the day, but am essentially a self taught amateur, so I'd very much appreciate answers that are at the undergrad physics major level and refer to a textbook which I can read up more on my own.

Please be pedantic, as the issues may be subtle. This started from a co-worker (who unlike me has a PhD in physics), who tried to dismiss a proposed theory of quantum gravity (draft here) based on some very basic arguments. The physics Professor suggesting the theory has publicly stated that my co-worker is completely ignoring the meat of the proposal because he is misunderstanding the basics of QFT and GR. They both are claiming the other is so wrong they should basically go back to school. So this has become like a fascinating physics debate to me. I know many times when learning a difficult subject that seemingly obvious conclusions from basic concepts can be wrong, and while my co-worker's arguments are very convincing to me, it is so basic that it does raise some worries that he is just misunderstanding the Professor. I kept asking questions, so my coworker sent me here for unbiased help.

The issue in question is the very starting point of the proposal which is to define a Hilbert space over some scalars other than the complex numbers. In particular, a subset of 4x4 matrices which can be written as $x_a \gamma^a$ where ($x_a$ are four real numbers, and $\gamma^a$ are the dirac matrices). The paper refers to $x_a$ as four-vectors in the algebra $C\ell_{1,3}(R)$ (the algebra of the dirac matrices). So the paper's starting point is a "Hilbert space over the space of four-vectors".

As I understand it, two simplified arguments against this theory are:

# First line of reasoning

A Hilbert space $H$ over some scalars $S$ must satisfy:

1. For any vector $X \in H$, and scalar $a \in S$, then $aX \in H$ (scalar multiplication gives another vector in $H$)

2. For any two vectors $X,Y \in H$, the inner product $\langle X|Y\rangle \in S$ (the inner product of two vectors is a scalar in the Hilbert space)

3. For any two vectors $X,Y \in H$, and scalar $a \in S$, the inner product $\langle X|aY\rangle = \langle X|Y \rangle a$ (the inner product is linear)

Then, by applying #1 and #2 $\langle X|aY\rangle \in S$. Then by applying this fact along with #3, for an arbitrary scalar $a \in S$, and any scalar which can be written as the result of an inner product $b=\langle X|Y\rangle$, the result of multiplying these two scalars in $S$, should also be a scalar in $S$ (simply, $ba \in S$).

Therefore, a counter-example to the existence of this Hilbert space, is to show that two four-vectors multiplied in the algebra $C\ell_{1,3}(R)$ is not a four-vector (in other words, the four-vectors are not a sub-algebra of $C\ell_{1,3}(R)$ ).

This can be broken down into simple matrix algebra (for example as my coworker did here) to show that multiplication of these "scalars" is not closed in the set of these scalars, because multiplying two four-vectors can yield something other than a four-vector. I have worked this out more generally, and it appears that multiplying any two four-vectors will not be another four-vector unless at least one four-vector is the zero vector (0,0,0,0). I do not fully trust my work here enough to discount a professor though. Is one counter-example enough? Or is it possible that once we restrict to just scalars which are the result of inner-products, that somehow the product is closed?

In a reference the professor recommended, the product in Clifford algebras is discussed fairly clearly (here). If I read that correctly, the product of any two vectors will be the sum of scalars and bivectors in the multivector space. Therefore, NO result of multiplying two four-vectors can be written as another four-vector except in the case that at least one four-vectors is the zero vector (0,0,0,0). This agrees with the messy matrix algebra I worked out, and makes me more confident. But other people have explained similarly, and the professor stated that they are misreading that source. Is there a better source? What is wrong with the above logic?

Again, this seems suspiciously simple, and if an amateur is disagreeing with a physics professor, its usually not a good sign. If it wasn't for my co-worker, I'd worry I'm becoming a crackpot. Am I missing something fundamental here?

# Second line of reasoning

In quantum mechanics:

1. States of a system are represented by vectors in the Hilbert space. (Although not uniquely, as vectors related by a scalar multiplication represent the same physical state.)

2. Observables are self-adjoint operators on the Hilbert space. Measuring an observable will place the system in an eigen vector of that operator.

3. For a system prepared in a state represented by $X$, the probability of measuring it to be in a state represented by $Y$ (some non-degenerate eigen vector of the observable being measured) is $$Prob = \frac{\langle X|Y\rangle \langle Y|X\rangle}{\langle X|X\rangle \langle Y|Y \rangle}$$

Therefore, to predict measurements, we also need to be able to divide with these scalars. And furthermore, the result of any calculation of the above form should be a real number for it to make sense as a probability. Yet $C\ell_{1,3}(R)$ is not a normed division algebra. And since $\langle X | X \rangle$ is not even real valued in the professor's theory, I don't understand how the predicted probabilities could be.

The professor's response to this (and possibly some of the properties of Hilbert spaces above as well? it is not clear to me) is that this is a quantum field theory and the properties of Hilbert spaces and calculating probabilities for measurements, work differently in QFT than in introductory non-relativistic particle QM.

I went to look this up in the books I have, and surprisingly in Srednicki "Quantum Field Theory", he explicitly states in the very first page of the first chapter that he will not be going over the postulates of QFT, and in the "preface for students" he just lists some equations and says if you understand those you have the background to use this book. He is just assuming we already know the postulates?

I see in this physics stackexchange question (formalism of quantum field theory vs quantum mechanics), that at least in Lubos's opinion this is because the postulates are the same. But he doesn't give any references, and the professor's suggested reference ("Axiomatic quantum field theory in curved spacetime" Hollands and Wald), doesn't even discuss measurements or probabilities. In searching, I found the oft cited "Postulates of Quantum Field Theory" Hagg and Schorer (1962), but even that doesn't discuss measurements or probabilities. They just discuss how to build up a Hilbert space for field theories. They seem to just assume we know the 'rest' of the postulates. I also have "Quantum Field Theory in a Nutshell" by Zee, I skimmed the beginning and he too seems to just assume we know the postulates.

If it wasn't for my co-worker, at this point I'd just assume I'm wrong, as I can't even find a book to validate my understanding of the postulates and I'm an amateur disagreeing with a currently employed physics Professor.

Can someone please help me evaluate this physics theory, and give me some textbook references I can follow up on?

• I suppose, but I am not sure, he is doing QFT over the field of quaternions. I don't know a lot on the subject, but I think you should look in that direction. Quaternions have a matrix $4\times 4$ representation, and are linked to Clifford algebras. – yuggib Sep 27 '14 at 6:10
• Quaternions form a different algebra, $Cl_{0,2}(R)$. While this does have a 4x4 real matrix representation, this is different from the algebra of the Dirac matrices. – CuriousKev Sep 27 '14 at 6:37
• @Qmechanic Thank you for trying to touch up the title, but the 'numbers' in question here are not number fields. There may be a more appropriate term, but it should not be fields, as even quaternions (discussed in both answers) are not a mathematical field. – CuriousKev Sep 27 '14 at 21:59
• An article on a related topic, which may interest the OP, Twisted de Rham cohomology, homological definition of the integral and "Physics over a ring"... – Alex Nelson Oct 3 '14 at 15:33

## 2 Answers

Actually, all that is quite known from the foundational work by von Neumann and Birkhoff. In this formulation of QM (and in the subsequent evolution of this research area) one constructs the quantum theory theory starting from the lattice of elementary "YES-NO" observables (see my answer on quantum probabilities for more details) or "elemetary propositions" experimentally testable on a quantum system. These assumptions describe the common phenomenology of all quantum systems. That lattice turns out to be $\sigma$-complete, orthomodular, separable, atomic, irreducible and verifying the so called "covering property". In the standard QM this lattice is the one of orthogonal projectors in a complex Hilbert space. However already von Neumann noticed that at least two other possibilities seemed feasible in principle: the lattice of orthogonal projectors in a real Hilbert space and the lattice of orthogonal projectors in a quaternionic Hilbert space. In all cases, states are generalized probability measures on the relevant lattice.

This idea remained a long standing conjecture till 1995, when Solér (e.g. have a look at this entry of Stanford Encyclopedia of Philosophy) proved von Neumann's conjecture ruling out other formulations on different Hilbert space-like structures (e.g. using Clifford algebras as space of scalars). It rules out also Hilbert spaces constructed on the non associative algebra of octonions in apparent contradiction to what stated in the paper you mention.

In the presence of a time reversal operation, real quantum mechanics can be proved to be equivalent to the standard complex version. Instead the quaternionic one could contain some new physics, at least it is the opinion of S. Adler who wrote a thick book on this idea from a very physical viewpoint.

Some of fundamental theoretical results in QM survive the passage to quaternionic QM, like Wigner, Kadison, Gleason theorems (the last one is fundamental as it proves that the states are nothing but densitiy matrices and vector states as assumed in more elementary formulations of QM). Varadarajan's book on the geometry of QM deals with the three formulations simultaneously.

Quaternionic QM involves an interesting noncommutative functional analysis form the pure mathematical viewpoint (see for instance this paper of mine).

For these reasons I do not think that the paper you mention - which assumes to deal with some sort of Hilbert space whose scalars are elements of a Clifford algebra - may present a theory consistent with the basic standard assumptions of quantum theories formulated in Hilbert spaces or generalizations. This is just in view of Sòler's theorem. However the paper is not written in that clear mathematical fashion as the subject would deserve, in my honest opinion, for so mathematically, physically, and philosophically delicate issues.

To be honest I might also say that Hilbert space formulation is not the only possible. A more recent and in a sense more powerful is the algebraic formulation where the fundamental objects are not elements of a lattice and generalized probability measures on that lattice, but they are elements of a unital $C^*$-algebra (or more weakly a $*$-algebra or a Jordan algebra) the Hermitian ones representing the observables of the system. States are now defined by normalized positive functionals on the algebra, representing expectation values. The celebrated GNS reconstruction theorem proves that, when a reference state is chosen, this algebraic picture is equivalent to a standard construction - à la von Neumann say - in a Hilbert space. There are however many unitarily inequivalent Hilbert space realizations of the same algebraic structure.

However, the paper you mention does not seem to deal with this more abstract formulation.

(Regarding Stefan Hollands-Bob Wald's paper, I know quite well the authors and the ideas contained in that paper and I discussed them with Stefan in the past. I cannot see well how these ideas have much to do with alternative formulations of quantum theories at the level of the question I am answering. The point, there, was the re-formulation of quantum field theory, avoiding the standard perturbative approach. As far as I remember the basic structure of the Hilbert space does not play any fundamental role.)

• Thank you for the explanations! Strangely enough the professor even mentions Soler's result and dismisses it: "Quantum Mechanical theories have to use an algebra of scalars which is isomorphic to one of these four R,C,H,O. (1) Quantum field theory however includes quantities which are four vectors in Minkowski space which leads to special considerations." Since you actually work in this field, would you mind commenting on the paper at TheWinnower? It could be useful to the author and others (most recent discussion is here: thewinnower.com/papers/fundamentals-of-relativization ) – CuriousKev Sep 27 '14 at 8:19
• I see, it is another viewpoint explicitly referring to QFT and not QM. Actually I do not know that paper you point out, I will have a look at it later. Thank you very much. – Valter Moretti Sep 27 '14 at 8:29
• Besides that offhand comment, the paper never discusses Soler's result and how it hopes to escape this. Since the paper is still try to use a Hilbert space for formulating QFT, can one just use the Soler result as a no-go theorem to say the very starting point of this idea cannot be fruitful? Are there any loop holes that you know of? – CuriousKev Sep 27 '14 at 21:50
• My question was about help evaluating the professor's paper. So I'm not sure how to interpret your last comment. In the answer, you seem to rule out this type of theory, but then in the last comment you say the paper is "another viewpoint"? But your answer seems to directly apply, how is this "another viewpoint"? As a starting point the professor is trying to define a Hilbert space over four-vectors, and as I feel I've shown above, this isn't even mathematically well defined. The professor's idea doesn't appear to be "another viewpoint", it just appears to be wrong. No? What am I missing? – CuriousKev Sep 28 '14 at 21:37
• By the way, the professor's paper is here: thewinnower.com/papers/… The link I gave in the last comment was just the professor's response to some reviewers on the paper. – CuriousKev Sep 28 '14 at 21:40

Ok, to expand my comment into an answer:

There are only two finite-dimensional division rings (that admit division) containing the real numbers as a finite subring: the complex numbers and the quaternions (application of Frobenius theorem). Also, a vector space (and Hilbert spaces are vector spaces) is usually defined over a field, that is a non-zero commutative ring. Over a ring you have the module that is a generalization of a vector space on non-commutative rings (as the quaternions, and the supposed Dirac matrix algebra).

So even supposing you can generalize a Hilbert space as a module over a ring with a scalar product (I am not sure it is possible); the only way to allow for division, and to interpret suitable scalar products as a real probability, is to use either reals, complex (usual Hilbert spaces), quaternions or infinite dimensional division rings. Since it does not seem to be the case here, I suppose: either he is referring to quaternions in some sense, or he is mistaken, or he has to redefine from scratch the concept of probabilities, Hilbert space etc...

• Good point that this is technically discussing a Module with an inner product. But because multiplication in $C\ell_{1,3}(R)$ is not closed for four-vectors, the professor doesn't even have a ring to define a module over, correct? I guess my point in 'first line of reasoning' is ultimately that one can't just take a subset of an algebra and assume you still have a closed algebra. Am I oversimplying here? This paper just seems so very wrong to me. – CuriousKev Sep 27 '14 at 21:33
• Like Valter Moretti above, you also mention reals, complex numbers, and quaternions. The Soler result he mentions seems to be a no-go theorem for anything else, but you also mention "infinite dimensional division rings", so is there a loop hole in the Soler result? More directly, can the Soler result be used as a no-go theorem to show that this paper's starting point cannot be fruitful? – CuriousKev Sep 27 '14 at 21:38
• mathoverflow.net/q/45653 concludes there are no infinite-dimensional normed division algebras. – CuriousKev Sep 27 '14 at 21:43
• yes, he does not seem to have a ring to start with, if the algebra is not closed. I agree it is not clear whether his system is meaningful from the beginning, but valter is surely more expert in the field than me ;-) – yuggib Sep 27 '14 at 22:49